Giải thích các bước giải:
a,
Ta có:
\(\begin{array}{l}
P = \left( {\frac{{a\sqrt a - 1}}{{a - \sqrt a }} - \frac{{a\sqrt a + 1}}{{a + \sqrt a }}} \right):\frac{{a + 2}}{{a - 2}}\\
= \left( {\frac{{\left( {\sqrt a - 1} \right)\left( {a + \sqrt a + 1} \right)}}{{\sqrt a \left( {\sqrt a - 1} \right)}} - \frac{{\left( {\sqrt a + 1} \right)\left( {a - \sqrt a + 1} \right)}}{{\sqrt a \left( {\sqrt a + 1} \right)}}} \right):\frac{{a + 2}}{{a - 2}}\\
= \left( {\frac{{a + \sqrt a + 1}}{{\sqrt a }} - \frac{{a - \sqrt a + 1}}{{\sqrt a }}} \right):\frac{{a + 2}}{{a - 2}}\\
= \frac{{2\sqrt a }}{{\sqrt a }}.\frac{{a - 2}}{{a + 2}}\\
= \frac{{2\left( {a - 2} \right)}}{{a + 2}}\\
b,\\
P = 2.\frac{{a - 2}}{{a + 2}} = 2.\left( {1 - \frac{4}{{a + 2}}} \right) = 2 - \frac{8}{{a + 2}}\\
P \in Z \Leftrightarrow \frac{8}{{a + 2}} \in Z\\
\Rightarrow a + 2 \in \left\{ { \pm 1;\,\,\, \pm 2;\,\, \pm 4;\,\,\, \pm 8} \right\}\\
\Rightarrow a \in \left\{ { - 10;\,\, - 6;\,\, - 4;\,\, - 3;\,\, - 1;\,\,0;\,\,2;\,\,6} \right\}\\
a > 0,\,\,a \ne 1;\,\,a \ne 2\\
\Rightarrow a = 6
\end{array}\)
